What is the solution for (dy)/(dx)+1/(x+1)y=x^{-2} ?

The solution for (dy)/(dx)+1/(x+1)y=x^{-2} is y=(ln(x))/(x+1)-1/(x(x+1))+(c_{1})/(x+1)

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(dy)/(dx)+1/(x+1)y=x^{-2}

\frac{d y}{d x} + \frac{1}{x + 1} y = x^{- 2}

y = \frac{ln ( x )}{x + 1} - \frac{1}{x ( x + 1 )} + \frac{c _{1}}{x + 1}

Solution steps

\frac{d y}{d x} + \frac{1}{x + 1} y = x^{- 2}

Solve linear ODE:

y = \frac{ln ( x )}{x + 1} - \frac{1}{x ( x + 1 )} + \frac{c _{1}}{x + 1}

y = \frac{ln ( x )}{x + 1} - \frac{1}{x ( x + 1 )} + \frac{c _{1}}{x + 1}

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e^{x} + e^{- x}

\frac{\partial}{\partial x} (cos (x + e^{y}))

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What is the solution for (dy)/(dx)+1/(x+1)y=x^{-2} ?
The solution for (dy)/(dx)+1/(x+1)y=x^{-2} is y=(ln(x))/(x+1)-1/(x(x+1))+(c_{1})/(x+1)